Here we apply the Clausius-Clapeyron equation:
ln(P₁/P₂) = ΔH/R x (1/T₂ - 1/T₁)
The normal vapor pressure is 4.24 kPa (P₁)
The boiling point at this pressure is 293 K (P₂)
The heat of vaporization is 39.9 kJ/mol (ΔH)
We need to find the vapor pressure (P₂) at the given temperature 355.3 K (T₂)
ln(4.24/P₂) = 39.9/0.008314 x (1/355.3 - 1/293)
P₂ = 101.2 kPa
<span>Halflife is the time needed for a radioactive molecule to decay half of its current mass. If t is the time elapsed, the formula for the halflife would be:
final mass= original mass * </span>

<span>
If you put the information of the problem into the formula, the equation will be:
</span>final mass= original mass *

final mass= 800g *
<span>A. the sum of the protons and neutrons in one atom of the element</span>
Answer:
The number of chlorine atoms present in
units of gold III chloride is
Explanation:
Formula of Gold (III) chloride: 
<em>Avogadro Number</em> : Number of particles present in <u>one mole</u><u> </u>of a substance.
Using,


= 1 mole(0.9999 , nearly equal to 1 )
The given Gold III chloride sample is 1 mole in amount.
= 1 mole of 
In this Sample,
1 mole of
will give = 3 mole of Chlorine atoms
1 mole of Cl contain =
3 mole of Cl contain = 
3 mole of Cl contain =
So,
The number of chlorine atoms present in
units of gold III chloride is